NCERT Solutions for Class 6 Maths Chapter 8 Decimals

# NCERT Solutions for Class 6 Maths Chapter 8 Decimals

Edited By Ramraj Saini | Updated on Nov 30, 2023 09:53 AM IST

NCERT Solutions for Class 6 Maths Chapter 8 Decimals are discussed here. Subject matter at Careers360 has created detailed solutions considering the latest syllabus of CBSE 2023. Students can check detailed NCERT Class 6 Maths solutions for chapter 8 in this article. There are many questions in the Class 6 solutions of NCERT Maths chapter 8 which will give more clarity to the concepts. In this NCERT 6 Maths solutions chapter 8 students will study about addition and subtraction of decimal numbers and other concepts.

There are 38 questions in 6 exercises of this chapter of NCERT. You will get detailed explanations of these questions in the CBSE NCERT solutions for Class 6 Maths chapter 8. Referring to NCERT Class 6 Maths solutions of chapter 8 helps to understand the chapters of NCERT Class 6 Syllabus in a more easy way. Topic-wise and exercise-wise solved questions are available below in the NCERT solutions for Class 6 Maths chapter 8 Decimals. You can also access NCERT Solutions for all Classes.

## NCERT Solutions for Class 6 Maths Chapter 8 Decimals - Important Formulae

1. Tenths = 1/10

2. Hundredths = 1/100

3. The decimal point is moved to the left by the number of zeros in the denominator

For example 150/100 = 1.5 (shifting decimal point by 2 points to left)

For more, Download Ebook - NCERT Class 6 Maths: Chapterwise Important Formulas And Points

## NCERT Solutions for Class 6 Maths Chapter 8 Decimals - Important Points

Every decimal can be written as a fraction: Decimals can be expressed as fractions, where the decimal part is written as the numerator and the place value of the decimal determines the denominator. For example, 0.5 can be written as 5/10 or 1/2.

Comparing decimal numbers: Decimal numbers can be compared by starting with the whole part. If the whole parts are equal, then the tenth parts can be compared, followed by the hundredth parts, and so on. By comparing the place values, you can determine which decimal number is greater or smaller.

Use of decimals: Decimals are used in various real-life situations, including money transactions, measurements, scores in sports, and scientific calculations. They provide a more precise representation of values compared to whole numbers.

Free download NCERT Solutions for Class 6 Maths Chapter 8 Decimals for CBSE Exam

## NCERT Solutions for Class 6 Maths Chapter 8 Decimals (Intext Questions and Exercise)

### Solutions of NCERT for Class 6 Maths Chapter 8 Decimals Topic 8.2 Tenths Question: 1 Can you now write the following as decimals

 Hundreds (100) Tens (10) Ones (1) Tenths $\small \left ( \frac{1}{10} \right )$ 5 3 8 1 2 7 3 4 3 5 4 6

Answer: Yes, we can write to them in Decimal form. The Numbers in the decimal forms are:

• i) 538.1
• ii) 273.4
• iii) 354.6

## Solutions of NCERT for Class 6 Maths Chapter 8 Decimals Topic 8.2 Subtopic Fractions as Decimals

Answer: The number in decimal forms are:

• $\frac{3}{2}=1.5$
• $\frac{4}{5}=0.8$
• $\frac{8}{5}=1.6$

## NCERT Solutions for Class 6 Maths Chapter 8 Decimals-Exercise: 8.1

Answer: Observing from the figure, we get

 Hundreds (100) Tens (10) Ones (1) Tenths $\small (\frac{1}{10})$ a) 0 3 1 2 b) 1 1 0 4

(a) $\small 19.4$ (b) $\small 0.3$ (c) $\small 10.6$ (d) $\small 205.9$

Answer: The Value table of the numbers are:

 Number Hundreds Tens Ones Tenths 19.3 0 1 9 3 0.3 0 0 0 3 10.6 0 1 0 6 205.9 2 0 5 9

Question: 3 Write each of the following as decimals :

(a) Seven-tenths (b) Two tens and nine-tenths

(c) Fourteen point six (d) One hundred and two ones

(e) Six hundred point eight

• (a) Seven-tenths :
• $7\times\frac{1}{10}=\frac{7}{10}=0.7$
• (b) Two tens and nine-tenths:
• $2\times10+9\times\frac{1}{10}=20+\frac{9}{10}=20+0.9=20.9$
• (c) Fourteen point six :
• $=14.6$
• (d) One hundred and two ones
• $1\times100+2\times1=100+2=102.0$
• (e) Six hundred point eight
• $=600.8$

Question: 4 Write each of the following as decimals:

(a) $\small \frac{5}{10}$ (b) $\small 3+\frac{7}{10}$ (c) $\small 200+60+5+\frac{1}{10}$ (d) $\small 70+\frac{8}{10}$
(e) $\small \frac{88}{10}$ (f) $\small 4\frac{2}{10}$ (g) $\small \frac{3}{2}$ (h) $\small \frac{2}{5}$ (i) $\small \frac{12}{5}$ (j) $\small 3\frac{3}{5}$ (k) $\small 4\frac{1}{2}$

Answer: As we know when we divide a number by 10, the result is that number with a decimal after one digit,(we count the digit from right to left). So Keeping that in mind,

• (a) $\small \frac{5}{10}=0.5$
• (b) $\small 3+\frac{7}{10}=3+0.7=3.7$
• (c) $\small 200+60+5+\frac{1}{10}=265+0.1=265.1$
• (d) $\small 70+\frac{8}{10}=70+0.8=70.8$
(e)
$\small \frac{88}{10}=8.8$
• (f) $\small 4\frac{2}{10}=4+\frac{2}{10}=4+0.2=4.2$
• (g) $\small \frac{3}{2}=\frac{3}{2}\times\frac{5}{5}=\frac{15}{10}=1.5$
• (h) $\small \frac{2}{5}=\frac{2}{5}\times\frac{2}{2}=\frac{4}{10}=0.4$
• (i) $\small \frac{12}{5}=\frac{12}{5}\times\frac{2}{2}=\frac{24}{10}=2.4$
• (j) $\small 3\frac{3}{5}=3+\frac{3}{5}=3+\frac{3}{5}\times\frac{2}{2}=3+\frac{6}{10}=3+0.6=3.6$
• (k) $\small 4\frac{1}{2}=4+\frac{1}{2}=4+\frac{1}{2}\times\frac{5}{5}=4+\frac{5}{10}=4+0.5=4.5$

(a) $\small 0.6$ (b) $\small 2.5$ (c) $\small 1.0$ (d) $\small 3.8$ (e) $\small 13.7$ (f) $\small 21.2$ (g) $\small 6.4$

Answer: Converting Decimals into Fractions. we get

• (a) $\small 0.6=\frac{6}{10}=\frac{3}{5}$
• (b) $\small 2.5=\frac{25}{10}=\frac{5}{2}$
• (c) $\small 1.0=1$
• (d) $\small 3.8=\frac{38}{10}=\frac{19}{5}$
• (e) $\small 13.7=\frac{137}{10}$
• (f) $\small 21.2=\frac{212}{10}=\frac{106}{5}$
• (g) $\small 6.4=\frac{64}{10}=\frac{32}{5}$

Question: 6 Express the following as cm using decimals.

(a) 2 mm (b) 30 mm (c) 116 mm (d) 4 cm 2 mm (e) 162 mm (f) 83 mm

1 cm = 10 mm
1 mm = 0.1 cm
So,

(a) 2 mm $=\frac{2}{10}\:cm=0.2\:cm$

(b) 30 mm $=\frac{30}{10}\:cm=3.0\: cm$

(c) 116 mm $=\frac{116}{10}\:cm=11.6\: cm$

(d) 4 cm 2 mm $=4\:cm+\frac{2}{10}\:cm=4+0.2=4.2\: cm$

(e) 162 mm $=\frac{162}{10}\:cm=16.2\: cm$

(f) 83 mm $=\frac{83}{10}\:cm=8.3\: cm$

(a) $\small 0.8$ (b) $\small 5.1$ (c) $\small 2.6$ (d) $\small 6.4$ (e) $\small 9.1$ (f) $\small 4.9$

• (a) $\small 0.8$ Lies Between 0 and 1. and 1 is the closest whole number to it.
• (b) $\small 5.1$ Lies between 5 and 6 and 5 is the closest whole number to it.
• (c) $\small 2.6$ Lies between 2 and 3 and 3 is the closest whole number to it.
• (d) $\small 6.4$ Lies between 6 and 7 and 6 is the closest whole number to it.
• (e) $\small 9.1$ Lies between 9 and 10 and 9 is the closest whole number to it.
• (f) $\small 4.9$ Lies between 4 and 5. and 5 is the closest whole number to it.

(a) $\small 0.2$ (b) $\small 1.9$ (c) $\small 1.1$ (d) $\small 2.5$

Answer: The numbers on the number line are:

(a) $\small 0.2$ (b) $\small 1.9$ (c) $\small 1.1$ (d) $\small 2.5$

Answer: As we can see the points lies in the number line,

$A=\frac{8}{10}=0.8$

$B=1+\frac{3}{10}=1+0.3=1.3$

$C=2+\frac{2}{10}=2+0.2=2.2$

$D=2+\frac{9}{10}=2+0.9=2.9$

Answer: length of Ramesh’s notebook = 9 cm 5 mm

As we know,

$1mm=\frac{1}{10}cm$

So Length in the unit of cm :

$=9cm+5\times\frac{1}{10}cm=9cm+0.5cm=9.5cm$

Hence the Length of Ramesh's notebook is 9.5 cm.

Answer: length of a young gram plant is 65 mm

As we know,

$1mm=\frac{1}{10}cm$

So,

$65mm=65\times\frac{1}{10}cm=\frac{65}{10}cm=6.5cm$

So Length of young gram plant is 6.5 cm.

## NCERT Solutions for Class 6 Maths Chapter 8 Decimals-Exercise: 8.2

 Ones Tenths Hundredsths Number (a) (b) (c)

 Ones Tenths Hundredths Number (a) 0 2 6 0.26 (b) 1 3 8 1.38 (c) 1 2 8 1.28

# Hundreds100Tens10Ones1TenthsHundredthsThousands(a)(b)(c)(d)(e)010200031132012260925320400521 Answer: As we can see from the table, the numbers are:

(a) $3+\frac{2}{10}+\frac{5}{100}=3+0.2+0.05=3.25$

(b) $100+2+\frac{6}{10}+\frac{3}{100}=100+2+0.6+0.03=102.63$

(c) $30+\frac{2}{100}+\frac{5}{1000}=30+0.02+0.005=30.025$

(d) $200+10+1+\frac{9}{10}+\frac{2}{1000}=211+0.9+0.002=211.902$

(e) $10+2+\frac{2}{10}+\frac{4}{100}+\frac{1}{1000}=12+0.2+0.04+0.001=12.241$

(a) $\small 0.29$ (b) $\small 2.08$ (c) $\small 19.60$ (d) $\small 148.32$ (e) $\small 200.812$

(a)

$\small 0.29=0.2+0.09=\frac{2}{10}+\frac{9}{100}$

(b)

$\small 2.08=2+0.08=2+\frac{8}{100}$

(c)

$\small 19.60=19+0.6=10+9+\frac{6}{10}$

(d)

$\small 148.32=148+0.3+0.02=100+40+8+\frac{3}{10}+\frac{2}{100}$

(e)

$\small 200.812=200+0.8+0.01+0.002=200+\frac{8}{10}+\frac{1}{100}+\frac{2}{1000}$

So, the table becomes:

 Hundreds Tens Ones Tenths Hundredths Thousands 0 0 0 2 9 0 0 0 2 0 8 0 0 1 9 6 0 0 1 4 8 3 2 0 2 0 0 8 1 2

Question: 4 Write each of the following as decimals.

(a) $\small 20+9+\frac{4}{10}+\frac{1}{100}$ (b) $\small 137+\frac{5}{100}$ (c) $\small \frac{7}{10}+\frac{6}{100}+\frac{4}{1000}$
(d) $\small 23+\frac{2}{10}+\frac{6}{1000}$ (e) $\small 700+20+5+\frac{9}{100}$

Writing the number in a decimal form:

(a)

$\small 20+9+\frac{4}{10}+\frac{1}{100}=29+0.4+0.01=29.41$

(b)

$\small 137+\frac{5}{100}=137+0.05=137.05$

(c)

$\small \frac{7}{10}+\frac{6}{100}+\frac{4}{1000}=0.7+0.06+0.004=0.764$
(d)

$\small 23+\frac{2}{10}+\frac{6}{1000}=23+0.2+0.006=23.206$

(e)

$\small 700+20+5+\frac{9}{100}=725+0.09=725.09$

(a) $\small 0.03$ (b) $\small 1.20$ (c) $\small 108.56$ (d) $\small 10.07$ (e) $\small 0.032$ (f) $\small 5.008$

Answer: As we know After decimal, we call one digit at a time.

• (a) $\small 0.03$ = Zero point zero three
• (b) $\small 1.20$ = One point two zero
• (c) $\small 108.56$ = One hundred eight point five six
• (d) $\small 10.07$ = Ten point zero seven
• (e) $\small 0.032$ = Zero point zero three two
• (f) $\small 5.008$ = Five point zero zero eight.

(a) $\small 0.06$ (b) $\small 0.45$ (c) $\small 0.19$ (d) $\small 0.66$ (e) $\small 0.92$ (f) $\small 0.57$

Answer: Here, we have a zoomed version of the number line in which the interval between two number is 0.1 instead of 1 which we use normally. So,

• (a) $\small 0.06$ lies between 0.0 and 0.1
• (b) $\small 0.45$ lies between 0.4 and 0.5
• (c) $\small 0.19$ lies between 0.1 and 0.2
• (d) $\small 0.66$ lies between 0.6 and 0.7
• (e) $\small 0.92$ lies between 0.9 and 1.0
• (f) $\small 0.57$ lies between 0.5 and 0.6.

Question:7 Write as fractions in lowest terms.

(a) $\small 0.60$ (b) $\small 0.05$ (c) $\small 0.75$ (d) $\small 0.18$ (e) $\small 0.25$ (f) $\small 0.125$ (g) $\small 0.066$

Answer: The Numbers in the lowest form of the fraction are:

(a)

$\small 0.60=\frac{60}{100}=\frac{6}{10}=\frac{3}{5}$

(b)

$\small 0.05=\frac{5}{100}=\frac{1}{20}$

(c)

$\small 0.75=\frac{75}{100}=\frac{3}{4}$

(d)

$\small 0.18=\frac{18}{100}=\frac{9}{50}$

(e)

$\small 0.25=\frac{25}{100}=\frac{1}{4}$

(f)

$\small 0.125=\frac{125}{1000}=\frac{5}{40}=\frac{1}{8}$

(g)

$\small 0.066=\frac{66}{1000}=\frac{33}{500}$

## CBSE NCERT Solutions for Class 6 Maths Chapter 8 Decimals-Exercise: 8.3

(g) $\small 1.5$ or $\small 1.50$ (h) $\small 1.431$ or $\small 1.490$ (i) $\small 3.3$ or $\small 3.300$ (j) $\small 5.64$ or $\small 5.603$

Answer: We compare Decimals just like the way we compare normal numbers. First, we see the leftmost digit of the numbers and if they are same then we move toward the right digits. So,

(a) $\small 0.3$ or $\small 0.4$

0.4 is greater as while number part (the number before the decimal) is the same.

(b) $\small 0.07$ or $\small 0.02$

0.07 is greater

(c) $\small 3$ or $\small 0.8$

3 is greater as the whole part of the numbers are 3 and 0.so obviously 3 is greater than 0.

(d) $\small 0.5$ or $\small 0.05$

0.5 is greater

(e) $\small 1.23$ or $\small 1.2$

1.23 is greater

(f) $\small 0.099$ or $\small 0.19$

0.19 is greater

(g) $\small 1.5$ or $\small 1.50$

As we know that after the decimal, we can add as many zeros as we want in the rightmost to the numbers just like we can add zeroes to the left side of any whole number. this we can do because those zeroes do not carry any value so they don't affect the numbers.

So. They both are the same number.

(h) $\small 1.431$ or $\small 1.490$

1.490 is greater

(i) $\small 3.3$ or $\small 3.300$

As we know in decimal the RIGHTMOST zeros don't carry any value.so,

They both are the same number

(j) $\small 5.64$ or $\small 5.603$

5.64 is greater.

Answer: we can take any two number from the number line and compare them. For example,

1) 9.9 > 3.3

2) 6.6 < 9.9

3) 3.6 > 3.3

4) 9.6 < 9.9

5) 6.9 > 3.9

## Solutions for NCERT Class 6 Maths Chapter 8 Decimals Topic 8.5 Using Decimals

1 rupee = 100 paise

$1\:paise=\frac{1}{100}\:rupees$

So

2 rupees 5 paise :

$=2+5\times\frac{1}{100}=2+\frac{5}{100}=2+0.05=2.05\:rupees$

and 2 rupees 50 paise:

$=2+50\times\frac{1}{100}=2+\frac{50}{100}=2+0.5=2.5\:rupees$

1 rupee = 100 paise

$1\:paise=\frac{1}{100}\:rupees$

So

20 rupees 7 paise:

$=20+7\times\frac{1}{100}=20+\frac{7}{100}=20+0.07=20.07\:rupees$

and 21 rupees 75 paise:

$=21+75\times\frac{1}{100}=21+\frac{75}{100}=21+0.75=21.75\:rupees$

## Solutions of NCERT for Class 6 Maths Chapter 8 Decimals Topic 8.5.2 Length

1 cm = 10 mm

$1 mm = \frac{1}{10}\:cm$

So,

$4 mm = 4\times\frac{1}{10}\:cm=\frac{4}{10}\:cm=0.4cm$ .

1 cm = 10 mm

$1mm=\frac{1}{10}cm$

So,

7cm 5 mm :

$=7 + 5\times\frac{1}{10}=7+\frac{5}{10}=7+0.5=7.5cm$ .

1 km = 1000 m

$1 m =\frac{1}{1000}km$

So,

52 m in km :

$52 m =52\times\frac{1}{1000}km=\frac{52}{1000}km=0.052km$
340 m in km :

$340m =340\times\frac{1}{1000}km=\frac{340}{1000}km=0.34km$

2008 m in km :

$2008m =2008\times\frac{1}{1000}km=\frac{2008}{1000}km=2.008km$

## Solutions of NCERT for Class 6 Maths Chapter 8 Decimals Topic 8.5.3 Weight

1 kg = 1000 g

$1\:g=\frac{1}{1000}\:kg$

So,

456 g in kg:

$456\:g=456\times\frac{1}{1000}\:kg=\frac{456}{1000}\:kg=0.456\:kg$

1 kg = 1000 g

$1\:g = \frac{1}{1000}\:kg$

So,

2 kg 9 g in kg:

$2\:kg+9\:g = 2\:kg+\frac{9}{1000}\:kg=2\:kg+0.009\:kg=2.009\:kg$

## NCERT Solutions for Class 6 Maths Chapter 8 Decimals-Exercise: 8.4

Question:1 Express as rupees using decimals.

(a) 5 paise (b) 75 paise (c) 20 paise (d) 50 rupees 90 paise (e) 725 paise

1 rupee = 100 paise

$1\:paise=\frac{1}{100}\:rupees$

So

(a) 5 paise

$=5\times\frac{1}{100}=\frac{5}{100}=0.05\:rupees$

(b) 75 paise

$=75\times\frac{1}{100}=\frac{75}{100}=0.75\:rupees$

(c) 20 paise

$=20\times\frac{1}{100}=\frac{20}{100}=0.20\:rupees$

(d) 50 rupees 90 paise

$=50+90\times\frac{1}{100}=50+\frac{90}{100}=50+0.9=50.9\:rupees$

(e) 725 paise

$=725\times\frac{1}{100}=\frac{725}{100}=7.25\:rupees$

Question:2 Express as meters using decimals.

(a) 15 cm (b) 6 cm (c) 2 m 45 cm (d) 9 m 7 cm (e) 419 cm

1 m = 100 cm

Which means

$1\:cm=\frac{1}{100}\:m$ .

So,

(a) 15 cm

$15\:cm=15\times\frac{1}{100}=\frac{15}{100}=0.15\:m$

(b) 6 cm

$6\:cm=6\times\frac{1}{100}=\frac{6}{100}=0.06\:m$

(c) 2 m 45 cm

$2+45\:cm=2+45\times\frac{1}{100}=2+\frac{45}{100}=2+0.45=2.45\:m$

(d) 9 m 7 cm

$9+7\:cm=9+7\times\frac{1}{100}=9+\frac{7}{100}=9+0.07=9.07\:m$

(e) 419 cm

$419\:cm=419\times\frac{1}{100}=\frac{419}{100}=4.19\:m$

Question:3 Express as cm using decimals.

(a) 5 mm (b) 60 mm (c) 164 mm (d) 9 cm 8 mm (e) 93 mm

1 cm = 10 mm

which means

$1\:mm=\frac{1}{10}\:cm$

So,

(a) 5 mm :

$5\:mm=5\times\frac{1}{10}=\frac{5}{10}=0.5\:cm$

(b) 60 mm

$60\:mm=60\times\frac{1}{10}=\frac{60}{10}=6.0\:cm$

(c) 164 mm

$164\:mm=164\times\frac{1}{10}=\frac{164}{10}=16.4\:cm$

(d) 9 cm 8 mm

$9\:cm+8\:mm=9+8\times\frac{1}{10}=9+\frac{8}{10}=9+0.8=9.8\:cm$

(e) 93 mm

$93\:mm=93\times\frac{1}{10}=\frac{93}{10}=9.3\:cm$

Question:4 Express as km using decimals.

(a) 8 m (b) 88 m (c) 8888 m (d) 70 km 5 m

1 km = 1000 m

which means

$1\:m=\frac{1}{1000}\:km$

So,

(a) 8 m

$8\:m=8\times\frac{1}{1000}=\frac{8}{1000}=0.008\:km$

(b) 88 m

$88\:m=88\times\frac{1}{1000}=\frac{88}{1000}=0.088\:km$

(c) 8888 m

$8888\:m=8888\times\frac{1}{1000}=\frac{8888}{1000}=8.888\:km$

(d) 70 km 5 m

$70\:km+5\:m=70+5\times\frac{1}{1000}=70+\frac{5}{1000}=70+0.005=70.005\:km$

Question:5 Express as kg using decimals.

(a) 2 g (b) 100 g (c) 3750 g
(d) 5 kg 8 g (e) 26 kg 50 g

1 kg = 1000 g

which means

$1 g = \frac{1}{1000}kg$

So,

(a) 2 g

$2 g = \frac{2}{1000}kg=0.002kg$

(b) 100 g

$100 g = \frac{100}{1000}kg=0.1kg$

(c) 3750 g

$3750 g = \frac{3750}{1000}kg=3.75kg$

(d) 5 kg 8 g

$5kg\:\:8g=5+8g =5kg+ \frac{8}{1000}kg=5kg+0.008kg=5.008kg$

(e) 26 kg 50 g

$26kg\:\:50g=26+50g =26kg+ \frac{50}{1000}kg=26kg+0.05kg=26.05kg$

## Solutions for NCERT Class 6 Maths Chapter 8 Decimals Topic 8.6 Addition of Numbers with Decimals

Question:(i) Find

$\small 0.29 + 0.36$

Answer: As we know that the decimal addition is exactly like normal addition and we don't change the position of decimal in the number.

So,

$\small \\0.29$

$\small +$ $\small \\0.36$

$\small \\0.65$

Hence

$\small 0.29 + 0.36=0.65$

Question:(ii) Find

$\small 0.7 + 0.08$

Answer: As we know that the decimal addition is exactly like normal addition and we don't change the position of decimal in the number.

So,

$0.70$

$+\:0.08$

$0.78$

Hence,

$\small 0.7 + 0.08=0.78$

Question:(iii) Find

$\small 1.54 + 1.80$

Answer: As we know that the decimal addition is exactly like normal addition and we don't change the position of decimal in the number.

So,

$1.54$

$+\:1.80$

$3.34$ .

Question:(iv) Find

$\small 2.66 + 1.85$

Answer: As we know that decimal addition is exactly like normal addition and we don't change the position of decimal in the number.

So,

$2.66$

$+\:1.85$

$4.51$

## Question:1 Find the sum in each of the following :

(a) $\small 0.007 + 8.5 + 30.08$

(b) $\small 15 + 0.632 + 13.8$

(c) $\small 27.076 + 0.55 + 0.004$

(d) $\small 25.65 + 9.005 + 3.7$

(e) $\small 0.75 + 10.425 + 2$

(f) $\small 280.69 + 25.2 + 38$

Answer: As we know that the decimal addition is exactly like normal addition and we don't change the position of decimal in the number.

So,

(a) $\small 0.007 + 8.5 + 30.08$

$0.007$

$8.500$

$+\:30.080$

$38.587$

(b) $\small 15 + 0.632 + 13.8$

$\small 15.000$

$\small 0.632$

$\small +\:13.800$

$\small 29.432$

(c) $\small 27.076 + 0.55 + 0.004$

$27.076$

$0.550$

$+\:0.004$

$27.630$

(d) $\small 25.65 + 9.005 + 3.7$

$25.650$

$9.005$

$+\:3.700$

$38.355$

(e) $\small 0.75 + 10.425 + 2$

$0.750$

$10.425$

$+\:2.000$

$13.175$

(f) $\small 280.69 + 25.2 + 38$

$280.69$

$25.20$

$+\:38.00$

$343.89$

Amount spend on maths book = Rs $\small 35.75$

Amount spend on science book = Rs $\small 32.60$

Total amount spend = Rs $\small 35.75$ $+$ Rs $\small 32.60$

= Rs $68.35$

Hence the total amount spends by Rashid is Rs 68.35.

The amount Radhika got from mother = Rs $\small 10.50$

The amount Radhika got from father = Rs $\small 15.80$

Total amount Radhika have = Rs $\small 10.50$ $+$ Rs $\small 15.80$

= Rs $26.30$

Hence the total amount of money Radhika got from her parents is Rs 26.30.

Length of cloth for shirt = 3 m 20 cm

Length of cloth for trouser = 2 m 5 cm

As we know.

For adding two numbers their unit must be the same.

So making length in meter:

$3m+20cm=3\:m+\frac{20}{100}m$

$=3\:m+0.2m$

$=3.2m$

and

$2m+5cm=2\:m+\frac{5}{100}m$

$=2\:m+0.05m$

$=2.05m$

Now,

Total Length of the cloth = 3.2 m + 2.05 m

= 5.25 m

Hence Nasreen bought clothe with a length of 5.25 m.

The distance covered in the morning = 2 km 35 m

The distance covered in the evening = 1 km 7 m

As we know.

For adding two numbers their unit must be the same.

So Lengths in the Kilometer :

$2 km+35m=2km+\frac{35}{1000}km$

$=2km+0.035km$

$=2.035km$

And

$1km+7m=1km+\frac{7}{1000}km$

$=1km+0.007km$

$=1.007\:km$

Now,

Total Distance covered by Naresh = 2.035 km + 1.007 km

= 3.042 km

Hence the distance covered by Naresh is 3.042 km.

The distance travelled by bus = 15 km 268 m

The distance travelled by car = 7 km 7 m

The distance travelled by foot = 500 m

As we know.

For adding two numbers their unit must be the same.

So,

$15 km +268 m = 15 km + \frac{268}{1000}km$

$= 15 km + 0.268km$

$= 15.268km$

Also

$7km+7m=7km+\frac{7}{1000}km$

$=7km+0.007km$

$=7.007km$

And,

$500m=\frac{500}{1000}km$

$=0.5km$

So,

The total distance Sunita traveled = 15.268 km + 7.007 km + 0.5 km

= 22.775 km

Hence the distance between Sunita's residence and school is 22.775 km.

amount of rice purchased by Ravi = 5 kg 400 g.

amount of sugar purchased by Ravi = 2 kg 20 g.

amount of flour purchased by Ravi = 10 kg 850 g.

As we know.

For adding two numbers their unit must be the same.

So,

$5\:kg+400g=5kg+\frac{400}{1000}kg$

$=5kg+0.4kg$

$=5.4kg$

Also,

$2\:kg+20g=2kg+\frac{20}{1000}kg$

$=2kg+0.02kg$

$=2.002kg$

And,

$10\:kg+850g=10kg+\frac{850}{1000}kg$

$=10kg+0.85kg$

$=10.85kg$

So,

The total weight of the grain Ravi purchased = 5.4 kg + 2.002 kg + 10.85 kg

= 18.270 kg

Hence Ravi purchased total grain having the weight of 18.270 kg.

## Question:1 Subtract $\small 1.85$ from $\small 5.46$

Answer: As we know, subtraction in decimal works exactly like normal subtraction, we just have to maintain the place of decimal.

So

$5.46$

$-1.85$

$3.61$

Answer: As we know, subtraction in decimal works exactly like normal subtraction, we just have to maintain the place of decimal.

So

$8.28$

$-5.25$

$3.03$

Answer: As we know, subtraction in decimal works exactly like normal subtraction, we just have to maintain the place of decimal.

So

$2.29$

$-0.95$

$1.34$

Answer: As we know, subtraction in decimal works exactly like normal subtraction, we just have to maintain the place of decimal.

So

$5.68$

$-2.25$

$3.43$

## CBSE NCERT Solutions for Class 7 Maths Chapter 8 Decimals-Exercise: 8.6

Question: 1 Subtract :

(a) Rs $\small 18.25$ from Rs $\small 20.75$

(b) $\small 202.54$ m from 250 m

(c) Rs $\small 5.36$ from Rs $\small 8.40$

(d) $\small 2.051$ km from $\small 5.206$ km

(e) $\small 0.314$ kg from $\small 2.107$ kg

Answer: As we know, subtraction in decimal works exactly like normal subtraction, we just have to maintain the place of decimal.

So

(a) Rs $\small 18.25$ from Rs $\small 20.75$

$20.75$

$-18.25$

$2.50$ Rs

(b) $\small 202.54$ m from 250 m\

$250.00$

$-202.54$

$47.46$ m

(c)Rs $\small 5.36$ from Rs $\small 8.40$

$8.40$

$-5.36$

$3.04$ Rs

(d ) $\small 2.051$ km from $\small 5.206$ km

$5.206$

$-2.051$

$3.155$ km

(e) $\small 0.314$ kg from $\small 2.107$ kg

$2.107$

$-0.314$

$1.793$ kg

Question: 2 Find the value of :

(a) $\small 9.756 - 6.28$

(b) $\small 21.05 -15.27$

(c) $\small 18.5 - 6.79$

(d) $\small 11.6 - 9.847$

Answer: As we know, subtraction in decimal works exactly like normal subtraction, we just have to maintain the place of decimal.

So

(a) $\small 9.756 - 6.28$

$9.756$

$-6.280$

$3.476$

(b) $\small 21.05 -15.27$

$21.05$

$-15.27$

$5.78$

(c) $\small 18.5 - 6.79$

$18.50$

$-6.79$

$11.71$

(d) $\small 11.6 - 9.847$

$11.600$

$-9.847$

$1.753$

Answer: Total money Raju gave to shopkeeper = Rs 50

The price of the book = Rs 35.65

Money shopkeeper will return Raju = Rs 50.00 - Rs 35.65

= Rs 14.35

Hence Raju got 14.35 Rs from the shopkeeper.

The money she spent on ice-cream = 11.75

The Remaining money Rani have = Rs 18.50 - Rs 11.75

= Rs 6.75

Hence Rani has 6.75 Rs now.

Answer: Length of cloth initially = 20 m 5 cm = 20.05 m

Length of cloth Tina cuts = 4 m 50 cm = 4.50 m

The length of remaining cloth = 20.05 m - 4.50 m

= 15.55 m

= 15 m 55 cm

Hence the length of remaining cloth Tina have is 15 m 55 cm.

Answer: Total distance travelled by Namita = 20 km 50 m = 20.05 km

The distance travelled by bus = 10 km 200 m = 10.20 km

The remaining distance which is traveled by auto = 20.05 km - 10.20 km

= 9.850 km.

Hence the distance travelled by Namita in auto is 9.850 km.

Answer: Total weight of the vegetables = 10 kg

weight of onion = 3 kg 500 g = 3.5 kg

weight of tomato = 2 kg 75 g = 2.075 kg

Total weight of onion and potato = 3.5 kg + 2.075 kg

= 5.575 kg

So, The rest weight of potato = 10 kg - 5.575 kg

= 4.425 kg

Hence, the weight of the potato is 4.425 kg.

## Decimals Class 6 Maths Chapter 8-Topics

• Tenths
• Hundredths
• Comparing Decimals
• Using Decimals
• Addition of Numbers with Decimals
• Subtraction of Decimals

## NCERT Solutions for Class 6 Mathematics - Chapter-Wise

 Chapters No. Chapters Name Chapter - 1 Chapter - 2 Chapter - 3 Chapter - 4 Chapter - 5 Chapter - 6 Chapter - 7 Chapter - 8 Decimals Chapter - 9 Chapter -10 Chapter -11 Chapter -12 Chapter -13 Chapter -14

## NCERT Solutions for Class 6 Mathematics - Subject Wise

 Solutions of NCERT for Class 6 Science

Some important findings from the NCERT solutions for Class 6 Maths chapter 8 Decimals which will be useful in real life too are listed below.

Converting a quantity from one unit to another-

• Millimeter to centimeter-

$1mm=\frac{1}{10}cm=0.1cm$

• Centimeter to meter-

$1cm=\frac{1}{100}m=0.01m$

• Paise to rupee

$1\ paise=\frac{1}{100}rupees=0.01 rupees$ .

$250\ paise=\frac{250}{100}rupees=2.5rupees$

• Grams to kg

$1gm=\frac{1}{1000}kg=0.001kg$

Tip- You don't have to remember these units conversion. If you know the value of the quantity in one unit and other units, you can convert it very easily. In CBSE NCERT solutions for class 6 maths chapter 8 decimals, you will find many unit conversion questions that will further strengthen your fundamentals.

Also Check NCERT Books and NCERT Syllabus here:

1. What are the topics Covered in NCERT solution Class 6 chapter 8?

The Topics are covered In NCERT given below

• Tenths
• Hundredths
• Comparing Decimals
• Using Decimals
• Addition of Numbers with Decimals
• Subtraction of Decimals
2. What are decimal numbers, and how are they represented?

Decimal numbers are used to represent numbers smaller than 1. The decimal point or period separates the whole number part from the fractional part in a decimal number. Each digit in a decimal number has a place value, which represents its value based on its position in the number. These concepts are important for exams, and you can find detailed explanations in the NCERT Class 6 Maths Solutions available in PDF format.

3. Will the class 6 maths chapter 8 decimals solutions help students understand important concepts for exams?

Yes, the NCERT Solutions for class 6 chapter 8 maths will indeed help students understand important concepts from an exam perspective. The solutions provide step-by-step explanations and examples that can assist students in grasping the expansion procedure and solving problems effectively. By practicing with the solutions PDF, students can become proficient in solving equations and achieve better scores in exams.

4. What are recurring and terminating decimals in the context of decimals covered in NCERT Solutions for maths class 6 chapter 8?

In decimals, a terminating decimal has a finite number of digits after the decimal point, and its value comes to an end. On the other hand, a recurring decimal keeps repeating the same sequence of digits infinitely after the decimal point. These concepts may appear in the question paper during the annual exam, and to solve them easily, students can refer to the NCERT Solutions and aim for good marks.

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